Re: Asteroid Problem

[Rick Tarara <rbtarara@SPRYNET.COM>]

I'm not as good at back of envelope calculations as Fermi ;-) and lost a
factor of ten, but the general principal is OK.  With the earth moving at 30
km/s along its orbital path, 15 minutes will move the earth 2 diameters.
To slow or speed up the arrival of the asteroid by 15 minutes rather than 2,
the asteroid's speed must be changed by .003% which is still only 2-3 m/s if
done 1 year before collision.  However, at masses of 10^15 kg or more, the
energy/force needed is still immense.

Rick

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Richard W. Tarara
Professor of Physics
Saint Mary's College
Notre Dame, IN 46556
219-284-4664
rtarara@saintmarys.edu

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----- Original Message -----
From: "Ludwik Kowalski" <kowalskiL@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Monday, August 27, 2001 10:12 PM
Subject: Re: Asteroid Problem


> Rick Tarara wrote:
>
> > For quick calculations, it might be easier to look at slowing down or
> > speeding up the asteroid.  While it depends on the angle of crossing
> > relative to the earth's path, I get something like a two minute change
in
> > arrival time as enough to allow the earth to clear the asteroid path.
This
> > is about a .0004% change (with 1 year lead time) which should require an
> > average speed change of less than 1 m/s (if the asteroid is moving at
less
> > than a quarter-million meters/s).   Plug in the mass and at least you
can
> > easily get the force/energy needed.
>
> 1 m/s over 2.5*10^5 m/s is 4*10-6. The corresponding
> relative change in kinetic energy is 16*10-12. Using
> your speed and m=10^19 kg (Michael said that asteroids
> are usually not as large as 10^20 kg) one gets the
> kinetic energy of 3*10^29 joules. The required change
> of kinetic energy must thus be equal to 16*10-12*3*10^29
> or only 4.8*10^18 joules.
>
> I would be more comfortable with the delay of two hours
> or more (instead of two minutes). This would increase the
> corresponding change in kinetic energy by many orders
> of magnitude.
>
> Compare the calculated above change in kinetic energy
> (work = 4.8*10^18 J) with the energy released by one very
> very large nuclear bomb (1000 megatons of TNT).
>
> 1 kg of TNT --> 5 MJ or 5*10^6 joules.
> 1000 megatons = 1,000,000,000 kg ---> 5*10^15 joules
>
> 4.8*10^18 / 5*10^15 = 960 bombs.
>
> And ten times more for a ten times more massive asteroid.
> Also note that one 1000 megaton bomb is like 100,000 of
> the Hiroshima-type bombs (0.01 megaton of TNT each).
> Keep in mind that this is for a delay of only two minutes.
> Ludwik Kowalski